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प्रश्न
उत्तर
\[LHS = \sin5x\]
\[ = \sin\left( 3x + 2x \right)\]
\[ = \sin3x \times \cos2x + \cos3x \times \sin2x\]
\[ = \left( 3\text{ sin } x - 4 \sin^3 x \right)\left( 2 \cos^2 x - 1 \right) + \left( 4 \cos^3 x - 3\text{ cos } x \right) \times 2\text{ sin } x \text{ cos } x\]
\[ = - 3\text{ sin } x + 4 \sin^3 x + 6\text{ sin } x \cos^2 x - 8 \sin^3 x \cos^2 x + 8\text{ sin } x \cos^4 x - 6\text{ sin } x \cos^2 x\]
\[ = 8\text{ sin } x \cos^4 x - 8 \sin^3 x \cos^2 x - 3\text{ sin } x + 4 \sin^3 x\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3\text{ sin } x + 3\text{ sin } x \cos^4 x + 4 \sin^3 x + 2 \sin^3 x \cos^2 x\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3\text{ sin} x\left( 1 - \cos^4 x \right) + 2 \sin^3 x\left( 2 + \cos^2 x \right)\]
\[= 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3\text{ sin }x\left( 1 - \cos^2 x \right)\left( 1 + \cos^2 x \right) + 2 \sin^3 x\left( 2 + \cos^2 x \right)\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - 3 \sin^3 x\left( 1 + \cos^2 x \right) + 2 \sin^3 x\left( 2 + \cos^2 x \right)\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x\left[ 3\left( 1 + \cos^2 x \right) - 2\left( 2 + \cos^2 x \right) \right]\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x\left[ 3 + 3 \cos^2 x - 4 - 2 \cos^2 x \right]\]
\[ = 5\text{ sin }x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x \left[ \cos^2 x - 1 \right]\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x - \sin^3 x \times \left( - \sin^2 x \right)\]
\[ = 5\text{ sin } x \cos^4 x - 10 \sin^3 x \cos^2 x + \sin^5 x\]
\[ = 5 \cos^4 x \text{ sin } x - 10 \cos^2 x \sin^3 x + \sin^5 x\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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